1.1 Polynomial Functions
Any function is a polynomial function if is a constant which belongs to the set of real numbers and the indices, are natural numbers. If then we say that is a polynomial of degree r.
Example1. is a polynomial of degree 4 and 1 is a zero of the polynomial as
Also,
2. is a polynomial of degree 3 and i is a zero of his polynomial as
Again,
3. is a polynomial of degree 2 and is a zero of this polynomial as
Note : The above definition and examples refer to polynomial functions in one variable. similarly polynomials in 2, 3, ..., n variables can be defined, the domain for polynomial in n variables being set of (ordered) n tuples of complex numbers and the range is the set of complex numbers.
Example : is a polynomial in of degree 2 as both and xy have degree 2 each.
Note : In a polynomial in n variables say a general term is where degree is where The degree of a polynomial in n variables is the maximum of the degrees of its terms.
Division in Polynomials
If and are any two polynomials then we can find polynomials and such hat where the degree of degree of
is called the quotient and the remainder.
In particular if is a polynomial with complex coefficients and a is a complex number then there exists a polynomial of degree 1 less than and a complex number R, such that
Example :
Here
and
1.3. Remainder Theorem and Factor Theorem
Reminder Theorem : If a polynomial is divided by then the remainder is equal to
Proof :…(1)
and so
If then and hence is a factor of
Further and thus a is a zero of the polynomial This leads to the factor theorem.
Factor Theorem : is a factor of polynomial if and only if
Fundamental theorem of algebra : Every polynomial function of degree has at least one zero in the complex numbers. In other words if we have
with then there exists at least one such that,
From this it is easy to deduce that a polynomial function of degree ‘n’ has exactly n zeroes.
Example 1.Find the polynomial function of lowest degree with integral coefficient s with as one of its zeroes.
Solution :Since the order of the surd is 2, you can expect the polynomial of the lowest degree to be a polynomial of degree 2.
Let
But is a zero, so and
and
So the required polynomial function is
You can find the other zero of this polynomial to i.e.,
Example 2.If x, y, z be positive numbers, show that
Solution :Since A.M. (arithmetic mean) (geometric men), therefore
Cubing both sides and multiplying throughout by 27, we have
Example 3.If be real numbers and none of the ’s be zero, then prove that
Solution :Applying Cauchy-Schwarz inequality to the numbers we have
or
Example 4.(Triangle Inequality). If be any real numbers, then show that
where the sign denotes the positive square root.
Solution :…(i)
By Cauchy-Schwarz inequality,
i.e., …(ii)
From (i) and (ii), we have
or
Taking positive square roots, we have
Remark : Geometrically interpreted, the above inequality expresses the fact the sum of two sides of a triangle can never be less than the third side and this is precisely the reason for the name ‘triangle inequality’.
Example 5.If be positive real numbers, show that
When does the inequality reduce to equality ?
Solution :If be real numbers, then by Cauchy-Schwarz inequality,
…(1)
Putting in the above inequality, we have
…(2)
Again, putting in (1), we have
…(3)
Squaring both sides of (3) and using (2), we immediately have
The above inequality reduces to an equality iff each of the inequalities (2) and (3) reduces to an equality, i.e., iff
and
i.e., iff
Example 6.If be positive real numbers such that then show that
Solution :Applying Cauchy-Schwarz inequality to the two sets of numbers
we have …(i)
Again, applying Cauchy-Schwarz inequality to the two sets of numbers
we have …(ii)
Squaring both sides of (i), we have
…(iii)
i.e.,
Since we have from (iii)
Taking positive square roots, we have
Tcheby Chef’s Inequality
Example 7.If are any real numbers such that then show that
Solution :Since therefore, are of the same sign or at least one of them is zero, so that
and therefore
Similarly, …(ii)
and …(iii)
Adding (i), (ii) and (iii) and then adding to both sides of resulting inequality, we have
Theorem : If and are any real numbers, such that
(i) then
(ii) then
Proof :(i)For every pair of distinct suffixes p and q, the differences and are of the same sign or at least one of them is zero.
Hence,
i.e.,
There are inequalities of the above type (for there are pairs of distinct suffixes p, q), Adding the corresponding sides of all such inequalities, we obtain
i.e.,
(ii) For every pair of distinct suffixes p and q, and are of opposite signs or at least one of them is zero. Hence,
i.e.,
Adding the corresponding sides of all the inequalities of the above type, we obtain
i.e.,
Remark : The inequality above can be put in the following symmetric form :
This form suggests the following generalisation which we state without proof.
If are real numbers such that
then,
We shall refer to this inequality as Generalised Tchebychef's Inequality.
Example 8.Show that :
(a)
(b)
Solution :(a)Applying Tchebychef's inequality to the sets of numbers
we have
or
or
Therefore,
(b)Applying Tchebychef's inequality to the sets of numbers 1,
we obtain
Taking positive square roots of both sides, we have
…(i)
Again, applying Tchebychef's inequality to the sets of numbers we have
…(ii)
From (i) and (ii), we have
Therefore
Example 9.If a, b, c are all positive and no two of them are equal, then prove that
(a)
(b)
Solution :(a)Without any loss of generality we may assume that By applying the generalised Tchebychef's inequality to three sets of numbers each of which is the same as we obtain
i.e., …(i)
Again, since the arithmetic mean exceeds the geometric mean
…(ii)
From (i) and (ii), we obtain the inequalities
…(a)
(b)As in (a), without any loss of generality we may assume that Since therefore,
Applying Tchebychef's inequality to the sets of numbers we obtain
…(iii)
Also, from (a) …(iv)
From (iii) and (iv), we have
Example 10.If a, b, c are positive and unequal, show that
Solution :
The differences are both of the same sign, and therefore, is positive. Similarly, the other two terms in the above sum are also positive. Therefore,
Example 11.If a, b, c are positive and if p, q, r are rational numbers such that and have the same sign, then show that
Show that if either
(i) or (ii) or (iii) then equality holds.
Solution :
Since and r have the same sign, the differences and have the same sign or are both zero.
Therefore,
and similarly each of the other two terms in the above sum is also non-negative, so that the sum is non-negative. This proves the inequality.
Also, if any of the given conditions is satisfied, then at least one of the factors in each term in vanishes and therefore the sum is zero. This proves that the equality holds.
ImportantTermsandResultsinAlgebra
1.Identities :
(a)If
(b)If
(c)If
2.Periodic function : A function f is said to be periodic, with period k..if.
3.Pigeon Hole Principle (PHP) : If more than n objects are distributed in ‘n’ boxes, then at least one box has more than one object in it.
4.Polynomials :
(a)A function f defined by
where is a positive integer or zero and are fixed complex numbers, is called a polynomial of degree n in x. The numbers are called the coefficients of f. If be a complex number such that then is said to be a zero of the polynomial f.
(b)If a polynomial is divided by where h is any complex number, the remainder is equal to f(h).
(c)If h is a zero of a polynomial , then is a factor of and conversely.
(d)Every polynomial equation of degree has exactly n roots.
(e)If a polynomial equation wish real coefficients has a complex root (p, q real numbers, ) then it also has a complex root
(f)If a polynomial equation with rational coefficients has an irrational root (p, q rational, q > 0, q not the square of a rational number), then it also has an irrational root
(g)If the rational number (a fraction in its lowest terms so that p, q are integers, prime to each other, ) is a root of the equation
where are integers and then p is a divisor of and is a divisor of .
(h)A number is a common root of the polynomial equations and iff it is a root of where is the G.C.D. of and .
(i)A number is a repeated root of a polynomial equation iff it is a common root of and
5.Functional equation : An equation involving an unknown function is called a functional equation.
6.(a)If be the roots of the equation then and .
(b)If be the roots of the equation then,
(c)If be the roots of the equations then,
Question 1.The product of two roots of the equation is 1, find all the roots.
Solution :Suppose the roots are and
Now,…(1)
…(2)
…(3)
…(4)
From Eq. (2) and Eq. (4), we get
…(5)
From Eq. (3) and Eq. (4), we get
…(6)
From Eq. (1) and Eq. (6), we get
or
Question 2.If are the roots of then prove that
(i)
(ii)
Solution :(i)Since are the roots of
…(1)
we have,
…(2)
From (2),
But from Eq. (1)
…(4)
Multiplying (1) by we get
…(5)
and are three roots of Eq. (5). So
…(6)
From Eq. (6),
…(7)
or
…(8)
Multiplying Eq. (1) by x, we get
…(9)
and hence
Again multiplying Eq. (1) by we get
…(10)
and hence
or
or
or
Question 3.Find the common roots of
and
hence solve the equations.
Solution :You can see that is H.C.F. of the two equations and hence, the common roots are the roots of
Now, …(1)
and …(2)
have 2 and 3 as their common roots.
If the other roots of Eq. (1) are and then
from eq. (1)
So, and are also roots of the quadratic equation
So the roots of Eq. (1) are
For Eq. (2), if and be the roots of Eq. 92), then we have
So and are the roots of
So the roots of Eq. (2) are
Question 4.Solve the system :
in terms of L.
Solution :Adding the three equations, we get
or
Dividing the three equations by we get
and solving we get,
and
Question 5.If and are non zero roots of the equation and respectively, prove that has a root between and
Solution : and are roots of
…(1)
and…(2)
respectively.
We have
and
Let
Thus,…(3)
…(4)
Adding in Eq. (3), we get
…(5)
Subtracting from Eq. (4), we get
Thus and have opposite signs and, hence, must have a root between and
Question 6.Find all real values of m such that both roots of the equation are greater than but less than +4.
Solution :The roots are i.e.,
gives
Question 7.The roots of the equation are in G.P. The sum of their reciprocal is 10. Compute the numerical value of |s|.
Solution :Let the roots be
…(1)
Sum of be reciprocals …(2)
Dividing (1) by (2), …(3)
Since s is the –ve of the product of the roots …(4)
or …(5)
Question 8.Let where a, b, c, d are constants. If
compute
Solution :We use a trick …(1)
The …(2)
i.e., divisible by …(3)
Since is a 4th degree polynomial
and…(4)
Question 9.Let be the polynomial equation of least possible degree with rational coefficients, having as a root, Compute the product of all the roots of
Solution :Let
i.e.,
Thus, and the product of the root is 56.
Question 10.The equations and have two roots in common. If the third root of each equation is represented by and respectively, compute the ordered pair
Solution :Common roots must be the roots of (Difference of equation)
Their sum is 0.
Then the third root of the first equation must be and of the second equation is –7.
Question 11.If are all real and and , find the value of
Solution :
Thus
where .
Question 12.If the integer A its reduced by the sum of its digits, the result is B. If B is increased by the sum of its digits, the result is A. Compute the largest 3-digt number A with this property.
Solution :A – (sum of the digits) must be divisible by 9. Then B + (sum of the digits) does not satisfy must be divisible by 9.
Now consider 999 :999 – 27 = 972(so defined sum of 27)
990 :990 – 18 = 972(so defined sum of 18)
Answer is 990.
Question 13.The roots of are If are real numbers, compute the minimum value of the sum
Solution :Sum of the roots = k; Sum of the roots taken two at a line = –k
Then
Thus …(1)
Thus minimum value of
Question 14.If then it must be true that for some integers a and b. Compute (a, b) where (b – a) as small as possible. Note : [x] represents the greatest integer function.
Solution :Replacing by y and solving,
which means
Ans. (–24, 18)
Question 15.The roots of are each one more than the roots of If are constants, compute
Solution :Now
Then
Question 16.Find all ordered pairs of positive integers (x, z) that
Solution :
Required ordered pairs are : (31, 29), (17, 13), (13, 7), (11, 1).
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